Science:Math Exam Resources/Courses/MATH152/April 2012/Question 01 (c)/Solution 1

From 1(b) we saw that the system of equations the line must satisfy is given by

${\begin{aligned}2x_{1}+x_{2}-x_{3}&=1\\x_{1}+x_{2}&=1.\end{aligned}}$

In order to find the parametric form we solve this system. First we put it in augmented form

$\left[{\begin{array}{ccc|c}2&1&-1&1\\1&1&0&1\end{array}}\right]$

and then perform row operations to reduce the matrix. First we will swap row 1 and row 2 so that the first pivot (row 1, column 1) has value 1.

$\left[{\begin{array}{ccc|c}1&1&0&1\\2&1&-1&1\end{array}}\right]$

and then we will subtract 2 multiples of row 1 from row 2 to place a zero below the pivot,

$\left[{\begin{array}{ccc|c}1&1&0&1\\0&-1&-1&-1\end{array}}\right].$

We will then multiply the second row by -1 to put a 1 in the pivot position there (row 2, column 2),

$\left[{\begin{array}{ccc|c}1&1&0&1\\0&1&1&1\end{array}}\right]$

and finally we will subtract 1 multiple of row 2 from row 1 to put a 0 above the pivot,

$\left[{\begin{array}{ccc|c}1&0&-1&0\\0&1&1&1\end{array}}\right].$

Notice that because we have more columns than rows, then we have a free variables (the matrix is rank deficient). Let the free variable be $x_{3}$, i.e. let $x_{3}=t$. Then from our reduced matrix problem we have that $x_{1}=t$ and that $x_{2}+t=1$ or $x_{2}=1-t$. Therefore we can write that

${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}0\\1\\0\end{bmatrix}}+t{\begin{bmatrix}1\\-1\\1\end{bmatrix}}.$

This parametrically describes the line.